本文介绍了使用php将数据插入表的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我的HTML表单如下:
My html form is like below:
<html>
<body>
<form action="createconnection.php" method="post">
Firstname : <input type="text", name="fname"> </br>
Lastname : <input type="test" name="lname"> </br>
Age : <input type="text" name="age"></br>
<input type="submit">
</form>
</body>
</html>
和php文件:
表名和数据库名都很好。
and php file:Table name and db name all are fine.
<?php
// Create connection
$con=mysqli_connect('127.0.0.1:3306' ,'root','root','my_db');
echo "hi";
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql="INSERT INTO table1 (Fname, LName, Age)
VALUES
('$_POST[fname]','$_POST[lname]','$_POST[age]')";
if (!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
}
echo "1 record added";
mysqli_close($con);
?>
我尝试了很多,但是当我执行它没有给出任何东西,没有在表中更新。
I tried very much but when I execute it gives nothing and no update in table.
推荐答案
这是一个非常简单的工作示例,您的代码与准备语句。
Here is a very simple working example of your code with prepared statements.
请注意查询上的询问符号, bind_param
, s
表示字符串, i
表示整数,
So ssi
means we will receive 2 strings and 1 integer entry.
<?php
// Your database info
$db_host = '';
$db_user = '';
$db_pass = '';
$db_name = '';
if (!empty($_POST))
{
$con = mysqli_connect($db_host, $db_user, $db_pass, $db_name);
if ($con->connect_error)
die('Connect Error (' . mysqli_connect_errno() . ') '. mysqli_connect_error());
$sql = "INSERT INTO table1 (Fname, LName, Age) VALUES (?,?,?)";
if (!$stmt = $con->prepare($sql))
die('Query failed: (' . $con->errno . ') ' . $con->error);
if (!$stmt->bind_param('ssi',$_POST['fname'],$_POST['lname'],$_POST['age']))
die('Bind Param failed: (' . $con->errno . ') ' . $con->error);
if (!$stmt->execute())
die('Insert Error ' . $con->error);
echo "Record added";
$stmt->close();
$con->close();
}
?>
<html>
<body>
<form action="createconnection.php" method="post">
Firstname : <input type="text", name="fname"> </br>
Lastname : <input type="test" name="lname"> </br>
Age : <input type="text" name="age"></br>
<input type="submit">
</form>
</body>
</html>
Just in case here is the SQL table used:
CREATE TABLE IF NOT EXISTS `table1` (
`Fname` varchar(50) NOT NULL,
`LName` varchar(50) NOT NULL,
`Age` int(3) NOT NULL
);
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